In a Young's double slit experiment, slits are seperated by a distance d and screen is at distance D, `(D gt gt d)` from slit plane. If light source of wavelength `lambda(lambda ltlt d)` is used. The minimum distance from central point on the screen where intensity is one fourth of the maximum is `(D lambda)/(nd)`. Find the value of n.

Answer

To solve the problem, we need to find the value of \( n \) in the expression for the minimum distance from the central point on the screen where the intensity is one fourth of the maximum intensity in a Young's double slit experiment.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two slits separated by a distance \( d \) and a screen at a distance \( D \) from the slit plane.
- The wavelength of light used is \( \lambda \), with the condition \( \lambda \ll d \).

2. **Intensity Formula**:
- The intensity \( I \) at a point on the screen due to two coherent sources (the slits) can be expressed as:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \]
- For equal intensities from both slits, let \( I_1 = I_2 = I_0 \). Thus, the formula simplifies to:
\[ I = 2I_0 (1 + \cos \phi) \]

3. **Condition for Intensity**:
- We are given that the intensity at the point of interest is one fourth of the maximum intensity:
\[ I = \frac{1}{4} I_{\text{max}} \]
- The maximum intensity \( I_{\text{max}} \) occurs at the central maximum where \( \phi = 0 \):
\[ I_{\text{max}} = 2I_0 (1 + 1) = 4I_0 \]
- Therefore, we have:
\[ I = \frac{1}{4} \times 4I_0 = I_0 \]

4. **Setting Up the Equation**:
- From the intensity equation:
\[ I_0 = 2I_0 (1 + \cos \phi) \]
- Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ 1 = 2(1 + \cos \phi) \]
- Rearranging gives:
\[ 1 = 2 + 2\cos \phi \implies 2\cos \phi = -1 \implies \cos \phi = -\frac{1}{2} \]

5. **Finding the Phase Difference**:
- The angle \( \phi \) corresponding to \( \cos \phi = -\frac{1}{2} \) is:
\[ \phi = \frac{2\pi}{3} \]

6. **Relating Phase Difference to Path Difference**:
- The phase difference \( \phi \) is related to the path difference \( \Delta x \) by:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
- Substituting for \( \phi \):
\[ \frac{2\pi}{3} = \frac{2\pi}{\lambda} \Delta x \]
- Canceling \( 2\pi \) gives:
\[ \Delta x = \frac{\lambda}{3} \]

7. **Using Geometry**:
- The path difference \( \Delta x \) can also be expressed in terms of the distance \( y \) from the central maximum on the screen:
\[ \Delta x = d \sin \theta \]
- For small angles, \( \sin \theta \approx \theta \approx \frac{y}{D} \):
\[ \Delta x = d \frac{y}{D} \]
- Setting the two expressions for \( \Delta x \) equal:
\[ \frac{\lambda}{3} = d \frac{y}{D} \]

8. **Solving for \( y \)**:
- Rearranging gives:
\[ y = \frac{\lambda D}{3d} \]

9. **Comparing with Given Expression**:
- The problem states that the minimum distance from the central point where intensity is one fourth of the maximum is given by:
\[ y = \frac{D \lambda}{nd} \]
- Comparing both expressions:
\[ \frac{\lambda D}{3d} = \frac{D \lambda}{nd} \]
- This implies:
\[ n = 3 \]

### Final Answer:
The value of \( n \) is \( 3 \).